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A colony of 100100 bacteria doubles in size every 6060 hours. What will the population be 180180 hours from now??

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Q. A colony of 100100 bacteria doubles in size every 6060 hours. What will the population be 180180 hours from now??
  1. Identify Parameters: Identify the initial population, the growth rate, and the total time for growth.\newlineInitial population P0P_0 = 100100 bacteria\newlineGrowth rate = doubles every 6060 hours\newlineTotal time tt = 180180 hours\newlineDetermine the number of doubling periods within the total time.
  2. Calculate Doubling Periods: Calculate the number of doubling periods.\newlineDoubling period TT = 6060 hours\newlineNumber of doubling periods nn = Total time / Doubling period\newlinen=180hours60hoursn = \frac{180 \, \text{hours}}{60 \, \text{hours}}\newlinen=3n = 3
  3. Apply Exponential Growth Formula: Apply the formula for exponential growth.\newlineFinal population PP = Initial population * (Growth factor)^Number of doubling periods\newlineGrowth factor for doubling = 22\newlineP=1002nP = 100 * 2^n\newlineP=10023P = 100 * 2^3
  4. Calculate Final Population: Calculate the final population.\newlineP = 100×23100 \times 2^3\newlineP = 100×8100 \times 8\newlineP = 800800

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